# Some nice polynomials related to Hankel determinants

Let $$f_n(x)=\prod_{j=0}^{\lfloor{\frac{n-1}{2}}\rfloor}\prod_{i=2j+1}^{2n-2j-1}\frac{2x+i}{i}$$ and $$g_n(x)=\prod_{j=1}^{\lfloor{\frac{n}{2}}\rfloor}\prod_{i=2j}^{2n-2j}\frac{2x+i}{i}.$$

Then $$f_n(k)=\det \left( {f_{n+i+j}(1) } \right)_{i,j = 0}^{k - 1}$$ for each positive integer $$k$$ and analogously for $$g_n(k).$$

Note that $$f_n(1)=\binom{2n+1}{n}$$ and $$g_n(1)=C_n,$$ a Catalan number.

Do these polynomials also occur in other contexts? Are there other integer sequences $$a_n$$

such that the polynomials $$p_n(x)$$ with $$p_n(k)=\det \left( {a_{n+i+j} } \right)_{i,j = 0}^{k - 1}$$ are nice?

• How do you define nice? Apr 8 at 19:17
• How do you define the products over $i$ when the lower index is larger than the upper index? Apr 8 at 21:00
• @ Richard Stanley:I have corrected the upper indices. Apr 9 at 7:26
• @ მამუკა ჯიბლაძე : such as above, only rational roots. Apr 9 at 7:29
• Is $g_n(x)$ perhaps the order polynomial of the (unshifted) staircase partition shape $(n,n-1,\ldots,1)$? As in doi.org/10.1016/S0195-6698(13)80128-X (or my previous question mathoverflow.net/questions/343005/…). Apr 9 at 12:27

See Example 4 in Section 3.1.6 of https://arxiv.org/abs/1409.2562 (Federico Ardila, "Algebraic and geometric methods in enumerative combinatorics").

It explains that your $$g_n(k)$$ has an interpretation in terms of nested fans of Dyck paths.

Equivalently, as I mentioned in a comment above, $$g_n(k)$$ is the order polynomial of the (unshifted) staircase partition $$(n,n-1,\ldots,1)$$ shape poset. In other words, it is the number of ways to fill the shape $$\begin{array}c \square & \square & \square & \square \\ \square & \square & \square \\ \square & \square \\ \square \end{array}$$ with nonnegative integers between $$0$$ and $$k$$ that are weakly decreassing in rows and columns.

I explained something similar in a previous MO question of yours.

EDIT: Since this question got bumped, let me edit it to mention also that there is also an analogous interpretation of your $$f_n(k)$$, in terms of the shifted double staircase shape poset, as explained in more detail in my comment: Generating functions for Hankel determinants of Catalan numbers.

• By the way, the product formula for the order polynomial is due to Proctor; see "New symmetric plane partitions identities..." (doi.org/10.1016/S0195-6698(13)80128-X) and "Odd symplectic groups" (doi.org/10.1007/BF01404455); although in fact he had an earlier proof based around manipulating a determinant of MacMahon. Apr 9 at 23:36
• Very interesting. Apr 10 at 13:39

This approach uses "Number Walls".

Given a sequence of elements $$\,a:\mathbb{Z}\to F\,$$ in a field $$\,F.\,$$ Define the infinite matrix of $$\,n\times n\,$$ Hankel determinants

$$W_{n,m} :=\det({a_{m-n+i+j}})_{i,j=0}^{n-1}. \tag{1}$$

According to a theorem on determinants by Jacobi, $$\,W_{n,m}\,$$ satisfies the Number Wall identity (except at $$n=m=0$$)

$$W_{n+1,m}W_{n-1,m} = W_{n,m+1}W_{n,m-1} - W_{n,m}W_{n,m} \tag{2}$$

if $$\,a_n=0\,$$ for all $$\,n<0.\,$$ The definition in equation $$(1)$$ implies

$$W_{0,m} = 1,\quad W_{1,m} = a_{m-1}. \tag{3}$$

Define the sequence of functions

$$p_n(k) := W_{k,n+k} = \det({a_{n+i+j}})_{i,j=0}^{k-1}. \tag{4}$$

Use this definition and equation $$(3)$$ to get

$$p_n(0) = 1, \quad p_n(1) = a_n. \tag{5}$$

For simplicity, suppose $$\,p_n\,$$ is a polynomial of degree $$\,n\,$$ for $$\,n=0,1.\,$$ Then equation $$(5)$$ implies

$$p_0(x) = a_0 = 1, \quad p_1(x) = x(a_1-1)+1. \tag{6}$$

The definition in equation $$(4)$$ implies $$\, p_0(k) = 1 = W_{k,k}\,$$ which is a linear equation in $$\,a_{2k}\,$$ which can be solved for when given values of all of the $$\,a_0,a_1,\dots,a_{2k-1}.\,$$

Similarly, $$\,p_1(k) = k(a_1-1)+1 = W_{k,k+1}\,$$ which is a linear equation in $$\,a_{2k+1}\,$$ which can be solved for when given values of all the $$\,a_0,a_1,\dots,a_{2k}.\,$$

Using this procedure, solve the equations successively to get (with $$\,z:=a_1$$)

$$a_1 \!=\! z, \;\; a_2 \!=\! 1+z^2\!, \;\; a_3 \!=\! 2+2z+z^3\!, \;\; a_4 \!=\! 6+4z+3z^2+z^4,\;\;\dots. \tag{7}$$

An alternative method is to rewrite the Number Wall identity equation $$(2)$$ as

$$p_{n+1}(k)\,p_{n-1}(k) = p_{n+1}(k-1)\,p_{n-1}(k+1) + p_n(k)\,p_n(k). \tag{8}$$

This implies the $$\,p_n(k)\,$$ recursion

$$p_n(k) = (p_n(k-1)\,p_{n-2}(k+1) + p_ {n-1}^2(k))/p_{n-2}(k). \tag{9}$$

Now use equation $$(5)$$ to get $$\,a_n = p_n(1).$$

The nice polynomial expression for $$\,a_n\,$$ is

$$a_n = \sum_{k=0}^n T_{n,k}\, z^k,\quad \frac2{1+2x+\sqrt{1-4x}-2xy} = \sum_{n,k} T_{n,k}\, x^n y^k \tag{10}$$

where $$\,T\,$$ is the triangular array in OEIS sequence A065600

Triangle T(n,k) giving number of Dyck paths of length 2n with exactly k hills (0 <= k <= n).

For $$\,z=0\,$$ we get OEIS sequence A000957 with a different offset.

For $$\,z=1\,$$ we get $$\,a_n=C_n.\,$$

For $$\,z=2\,$$ we get $$\,a_n=C_{n+1}.\,$$

For $$\,z=3\,$$ we get $$\,a_n=\binom{2n+1}{n}.\,$$

For $$\,z=4\,$$ we get OEIS sequence A049027.

All of these sequences with $$\,z>0\,$$ are the rows of the infinite array in OEIS sequence A076037 (with the first column of all 1s removed).

Note that $$p_2(x) = \frac16(1+x)((6-5x+2x^2)-(4x-4x^2)z+(x+2x^2)z^2). \tag{11}$$ This seems to factor into linear factors only if $$\,z=1,2,3.$$ $$p_3(x) = \frac1{180}(1+x)(2+x)(3+2x)F_3(x,z) \tag{12}$$ where $$\,F_3(x,z)\,$$ is a cubic in $$\,x\,$$ and $$\,z\,$$ which also seems to factor into linear factors only if $$\,z=1,2,3.$$ $$p_4(x) = \frac1{75600}(1+x)(2+x)^2(3+x)(3+2x)(5+2x)F_4(x,z) \tag{13}$$ where $$\,F_4(x,z)\,$$ is degree $$4$$ in $$\,x\,$$ and $$\,z.\,$$ It seems to factor into linear factors only if $$\,z=1,2,3.$$

This is consistent with the definitions of $$\,f_n(x)\,$$ and $$\,g_n(x)\,$$ given in the question.

• Is your equation (2) related to the octahedron recurrence? Apr 10 at 16:25
• @SamHopkins Yes, it is related. Also my MIT 2000 talk "Number Walls in Combinatorics". Also generalized Somos-4 sequences and elliptic divisibility sequences and several other topics. Apr 10 at 16:29
• @Somos: Thank you, this gives an answer to my second question. Unfortunately I cannot accept more than one answer. It turns out that the above examples are the only ones where all factors over the rationals are linear. In the general case it seems that setting $a_1=b$ then for $b>1$ the corresponding polynomials $P_n(b;x)$ can be expressed as $P_n(b;x)=\frac{1}{b-1}\det(\binom{x+i+j}{2j}+(b-2) \binom{x+i+j-1}{2j})$ with $0\leq i,j \leq{n-1}.$ Apr 11 at 14:14